# Does ζ(3)\zeta(3) have a connection with π\pi?

The problem

Can be $$ζ(3)\zeta(3)$$ written as $$απβ\alpha\pi^\beta$$, where ($$α,β∈C\alpha,\beta \in \mathbb{C}$$), $$β≠0\beta \ne 0$$ and $$α\alpha$$ doesn’t depend of $$π\pi$$ (like $$√2\sqrt2$$, for example)?

Details

Several $$ζ\zeta$$ values are connected with $$π\pi$$, like:

$$ζ\zeta$$(2)=$$π2/6\pi^2/6$$

$$ζ\zeta$$(4)=$$π4/90\pi^4/90$$

$$ζ\zeta$$(6)=$$π6/945\pi^6/945$$

and so on for all even numbers.

See this mathworld link to more details: Riemann Zeta Function

So the question is, could $$ζ(3)\zeta(3)$$ be written as:

$$ζ(3)=απβ\zeta(3)=\alpha\pi^\beta$$
$$α,β∈C\alpha,\beta \in \mathbb{C}$$
$$β≠0\beta \ne 0$$
$$α not dependent of π\alpha \text{ not dependent of } \pi$$

See $$α\alpha$$ not essencially belongs $$Q\mathbb{Q}$$ and $$α,β\alpha,\beta$$ could be real numbers too.

When I wrote $$α\alpha$$ is not dependent of $$π\pi$$ it’s a strange and a hard thing to be defined, but maybe $$α\alpha$$ can be written using $$ee$$ or $$γ\gamma$$ or $$√2\sqrt2$$ or some other constant.

Edit:

Maybe this still a open question. If

$$∑2k=0(−1)kB2k B2−2k+2(2k)! (2−2k+2)! \sum_{k = 0}^{2} (-1)^{k} \frac{B_{2k} \ B_{2- 2k + 2}}{(2k)! \ (2 - 2k + 2)!}$$

in $$−4∑2k=0(−1)kB2k B2−2k+2(2k)! (2−2k+2)!π3-4 \sum_{k = 0}^{2} (-1)^{k} \frac{B_{2k} \ B_{2- 2k + 2}}{(2k)! \ (2 - 2k + 2)!}\pi^3$$ be of the form $$δπ3\frac{\delta}{\pi^3}$$ with $$δ\delta$$ not dependent of $$π\pi$$

and $$−2∑k≥1k−3e2πk−1- 2 \sum_{k \geq 1} \frac{k^{-3}}{e^{2 \pi k} - 1}$$ not dependent of $$π\pi$$ too, this question still hard and open.

Edit 2:

I discovered a result, but later I’ve seen that this is something already known, either way, it is an interesting one to have it here.

$$ζ(3)=−4π2ζ′(−2)\zeta(3)=-4\pi^2\zeta'(-2)$$

but, if $$ζ′(−2)\zeta'(-2)$$ is of the form $$απ2\frac{\alpha}{\pi^2}$$, with $$α\alpha$$ not dependent of $$π\pi$$, then this still remains as a hard and an open question.

I have a conjecture that $$ζ′(−2)\zeta'(-2)$$ will not cancel the $$π2\pi^2$$ term, but since I wasn’t able to prove it and I can’t use it here.

We can express $$ζ\zeta$$ of odd numbers with $$ζ′\zeta'$$ in a easy way, with a “closed” form like this one.

The question is whether or not $\zeta(3)$ is connected with $\pi$. The answer is yes. Moreover, $\zeta(3) = \beta \pi^{\alpha}$ for some complex $\alpha, \beta$. Take $\alpha = 3$ and $\beta = 0.0387682...$. It is not known whether $\beta = 0.0387682...$ is algebraic or transcendental. It is known, however, that $\zeta(3)$ is irrational as shown by Apery.

Ramanujan conjectured and Grosswald proved that the following holds. If $\alpha, \beta > 0$ such that $\alpha \beta = \pi^{2}$, then for each non-negative integer $n$,

where $B_n$ is the $n^{\text{th}}$-Bernoulli number.

For odd positive integer $n$, we take $\alpha = \beta = \pi$,

In particular, for $n = 1$,

Observe that the coefficient of $\pi^{3}$ is rational, however, nothing is known about the algebraic nature of the infinite sum. This is a current topic of research. Indeed, it is conjectured that $\frac{\zeta(3)}{\pi^{3}}$ is transcendental.

Update: Recently, Takaaki Musha claims to have proved that $\frac{\zeta(2n+1)}{(2 \pi)^{2n+1}}$ is irrational for positive $n \geq 1$. However, some objection has since been raised (read comments below).